Question

Find the principal value of ${\sin ^{ - 1}}\left( {\sin \dfrac{{3{{\pi }}}}{5}} \right)$.

Answer 1

Hint: The principal value of an inverse trigonometric function at a point x is the value of the inverse function at the point x, which lies in the range of the principal branch. The principal value of the sine function is from $\left[ { - \dfrac{{{\pi }}}{2},\dfrac{{{\pi }}}{2}} \right]$.We will first bring the value of the variable within the principal range by applying the formula : $sin(180^o - A) = sinA$ and then cancel the inverse and general trigonometric functions to get our final answer.

Complete step-by-step answer:
In the given question we need to find the principal value of ${\sin ^{ - 1}}\left( {\sin \dfrac{{3{{\pi }}}}{5}} \right)$. We know that for sine function to be negative, the angle should be negative, that is less than $0^o$.

We know that ${{\pi }}$ rad = $180^o$, so
${\sin ^{ - 1}}\left( {\sin \dfrac{{3{{\pi }}}}{5}} \right) = sin^{-1}[sin108^o]$

We know that $sinA = sin(180^o - A)$
$sin^{-1}[sin108^o] = sin^{-1}[sin(180 - 72)^o] = sin^{-1}[sin(72)^o]$

By using $sin^{-1}[sinA] = A$, where $-90^o < A < 90^o$
$sin^{-1}[sin72^o] = 72^o$
=$ \dfrac{{2{{\pi }}}}{5}$
This is the required value.

Note: In such types of questions, we need to strictly follow the range of the principal values that have been specified. The principal values of the sine function here ranges from $-90^o$ to $90^o$. This is because there can be infinite values of any inverse trigonometric functions. A common mistake is that students directly apply the formula $sin^{-1}(sin(x)) = x$. This form of the formula is only applicable when x is an acute angle.